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AlekseyPX
3 years ago
9

If you had an aqueous mixture that contained Ag+ , K+ , and Pb2+ cations, how many different solids could precipitate if a chlor

ide solution was added?
Chemistry
1 answer:
SCORPION-xisa [38]3 years ago
3 0

Answer:

Two, KCl and PbCl₂.

Explanation:

Hello!

In this case, since the addition of chloride ions promote the following three ionic reactions:

Ag^+(aq)+Cl^-(aq)\rightleftharpoons AgCl(s)\\\\K^+(aq)+Cl^-(aq)\rightleftharpoons KCl(aq)\\\\Pb^{2+}(aq)+Cl^-(aq)\rightleftharpoons PbCl_2(s)

We can infer that both silver chloride and lead (II) chloride are precipitated products as their Ksp are 6.56x10⁻⁴ and 1.59x10⁻⁵ respectively, which means they are merely soluble in water.

Best regards!

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if 15 mol C is mixed with 10 mol O2 which reactant is the limiting reactant? Which reactant would be the excess reactant?
ioda

Answer:

C is the excess reactant.

Explanation:

Reaction is C + O2 --> CO2

1mol of C required to react with 1mol O2

Therefore 15 - 10 = 5moles of C will be in excess

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3 years ago
When 500.0 g of water is decomposed by electrolysis and the yield of hydrogen is only 75.3%, how much hydrogen chloride can be m
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The amount of hydrogen chloride that can be made is 1064 g

Why?

The two reactions are:

2H₂O → 2H₂ + O₂ 75.3 % yield

H₂ + Cl₂ → 2HCl 69.8% yield

We have to apply a big conversion factor to go from grams of water (The limiting reactant), to grams of HCl, the final product. We have to be very careful with the coefficients and percentage yields!

500.0gH_2O*\frac{1moleH_2O}{18.01 gH_2O}*\frac{2 moles H_2}{2 moles H_2O}*\frac{2.015g H_2}{1 mole H_2}*\frac{75.3 actual g}{100 theoretical g}=42.12 g H_2

42.12H_2*\frac{1 mole H_2}{2.015gH_2}*\frac{2 moles HCl}{1 mole H_2}*\frac{36.46g}{1 mole HCl}*\frac{69.8 actualg}{100 theoreticalg} =1064gHCl

Have a nice day!

#LearnwithBrainly

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